YES Termination w.r.t. Q proof of Transformed_CSR_04_Ex5_DLMMU04_GM.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSRRRProof (⇔, 80 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 26 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 MRRProof (⇔, 0 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 QDP
↳13 MRRProof (⇔, 7 ms)
↳14 QDP
↳15 DependencyGraphProof (⇔, 0 ms)
↳16 QDP
↳17 QDPOrderProof (⇔, 11 ms)
↳18 QDP
↳19 DependencyGraphProof (⇔, 0 ms)
↳20 QDP
↳21 UsableRulesProof (⇔, 0 ms)
↳22 QDP
↳23 QDPSizeChangeProof (⇔, 0 ms)
↳24 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(0, XS) → nil
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(nil, XS) → nil
a__zip(X, nil) → nil
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(nil) → nil
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__incr(x1)) = x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2 + 2·x1   
POL(a__tail(x1)) = 2·x1   
POL(a__take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(a__zip(x1, x2)) = 2 + 2·x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 1   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 2·x1   
POL(take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + x2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__take(0, XS) → nil
a__zip(nil, XS) → nil
a__zip(X, nil) → nil
a__repItems(nil) → nil


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__tail(cons(X, XS)) → mark(XS)
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2·x1   
POL(a__tail(x1)) = 2 + x1   
POL(a__take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(a__zip(x1, x2)) = 2 + 2·x1 + x2   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = 2·x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2·x1   
POL(s(x1)) = 2·x1   
POL(tail(x1)) = 2 + x1   
POL(take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + x2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

a__tail(cons(X, XS)) → mark(XS)


(4) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A__ODDNS → A__INCR(a__pairNs)
A__ODDNS → A__PAIRNS
A__INCR(cons(X, XS)) → MARK(X)
A__TAKE(s(N), cons(X, XS)) → MARK(X)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(pairNs) → A__PAIRNS
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(tail(X)) → A__TAIL(mark(X))
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
MARK(repItems(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__ODDNS → A__INCR(a__pairNs)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(tail(X)) → MARK(X)
MARK(repItems(X)) → A__REPITEMS(mark(X))
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A__TAKE(s(N), cons(X, XS)) → MARK(X)
MARK(take(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X2)
MARK(zip(X1, X2)) → A__ZIP(mark(X1), mark(X2))
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(X)
MARK(zip(X1, X2)) → MARK(X1)
MARK(zip(X1, X2)) → MARK(X2)
MARK(tail(X)) → MARK(X)
A__REPITEMS(cons(X, XS)) → MARK(X)
MARK(repItems(X)) → MARK(X)
A__ZIP(cons(X, XS), cons(Y, YS)) → MARK(Y)


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__INCR(x1)) = x1   
POL(A__ODDNS) = 0   
POL(A__REPITEMS(x1)) = 2 + x1   
POL(A__TAKE(x1, x2)) = 2 + x1 + x2   
POL(A__ZIP(x1, x2)) = 1 + x1 + x2   
POL(MARK(x1)) = x1   
POL(a__incr(x1)) = 2·x1   
POL(a__oddNs) = 0   
POL(a__pairNs) = 0   
POL(a__repItems(x1)) = 2 + 2·x1   
POL(a__tail(x1)) = 1 + x1   
POL(a__take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(a__zip(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = 2·x1   
POL(mark(x1)) = x1   
POL(nil) = 0   
POL(oddNs) = 0   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 0   
POL(repItems(x1)) = 2 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + x1   
POL(take(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(zip(x1, x2)) = 2 + 2·x1 + 2·x2   

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__ODDNS → A__INCR(a__pairNs)
MARK(take(X1, X2)) → A__TAKE(mark(X1), mark(X2))
MARK(repItems(X)) → A__REPITEMS(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(oddNs) → A__ODDNS
A__ODDNS → A__INCR(a__pairNs)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MARK(oddNs) → A__ODDNS


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__INCR(x1)) = 1 + x1   
POL(A__ODDNS) = 2   
POL(MARK(x1)) = 1 + 2·x1   
POL(a__incr(x1)) = x1   
POL(a__oddNs) = 1   
POL(a__pairNs) = 1   
POL(a__repItems(x1)) = 1 + 2·x1   
POL(a__tail(x1)) = x1   
POL(a__take(x1, x2)) = 1 + x1 + x2   
POL(a__zip(x1, x2)) = 1 + x1 + x2   
POL(cons(x1, x2)) = 2·x1 + x2   
POL(incr(x1)) = x1   
POL(mark(x1)) = x1   
POL(nil) = 2   
POL(oddNs) = 1   
POL(pair(x1, x2)) = x1 + x2   
POL(pairNs) = 1   
POL(repItems(x1)) = 1 + 2·x1   
POL(s(x1)) = x1   
POL(tail(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(zip(x1, x2)) = 1 + x1 + x2   

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(incr(X)) → A__INCR(mark(X))
A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__ODDNS → A__INCR(a__pairNs)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

A__INCR(cons(X, XS)) → MARK(X)
MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


A__INCR(cons(X, XS)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(s(X)) → MARK(X)
MARK(pair(X1, X2)) → MARK(X1)
MARK(pair(X1, X2)) → MARK(X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( A__INCR(x1) ) = 2x1 + 1

POL( mark(x1) ) = x1

POL( pairNs ) = 1

POL( a__pairNs ) = 1

POL( incr(x1) ) = 2x1

POL( a__incr(x1) ) = 2x1

POL( oddNs ) = 2

POL( a__oddNs ) = 2

POL( take(x1, x2) ) = 2x1 + 2x2 + 2

POL( a__take(x1, x2) ) = 2x1 + 2x2 + 2

POL( zip(x1, x2) ) = x1 + x2 + 2

POL( a__zip(x1, x2) ) = x1 + x2 + 2

POL( tail(x1) ) = 1

POL( a__tail(x1) ) = 1

POL( repItems(x1) ) = x1

POL( a__repItems(x1) ) = x1

POL( cons(x1, x2) ) = x1 + 1

POL( 0 ) = 0

POL( s(x1) ) = x1 + 1

POL( nil ) = 0

POL( pair(x1, x2) ) = x1 + x2 + 1

POL( MARK(x1) ) = x1 + 1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__incr(X) → incr(X)
a__oddNs → oddNs
a__oddNs → a__incr(a__pairNs)
a__pairNs → cons(0, incr(oddNs))
a__pairNs → pairNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(incr(X)) → A__INCR(mark(X))
MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(incr(X)) → MARK(X)

The TRS R consists of the following rules:

a__pairNs → cons(0, incr(oddNs))
a__oddNs → a__incr(a__pairNs)
a__incr(cons(X, XS)) → cons(s(mark(X)), incr(XS))
a__take(s(N), cons(X, XS)) → cons(mark(X), take(N, XS))
a__zip(cons(X, XS), cons(Y, YS)) → cons(pair(mark(X), mark(Y)), zip(XS, YS))
a__repItems(cons(X, XS)) → cons(mark(X), cons(X, repItems(XS)))
mark(pairNs) → a__pairNs
mark(incr(X)) → a__incr(mark(X))
mark(oddNs) → a__oddNs
mark(take(X1, X2)) → a__take(mark(X1), mark(X2))
mark(zip(X1, X2)) → a__zip(mark(X1), mark(X2))
mark(tail(X)) → a__tail(mark(X))
mark(repItems(X)) → a__repItems(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(nil) → nil
mark(pair(X1, X2)) → pair(mark(X1), mark(X2))
a__pairNs → pairNs
a__incr(X) → incr(X)
a__oddNs → oddNs
a__take(X1, X2) → take(X1, X2)
a__zip(X1, X2) → zip(X1, X2)
a__tail(X) → tail(X)
a__repItems(X) → repItems(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(incr(X)) → MARK(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MARK(incr(X)) → MARK(X)
    The graph contains the following edges 1 > 1

(24) YES